I am interested in criteria that guarantee that a Borel probability measure has compact support.
I outline two below and I am hoping to gather more as answers (if they exist).


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The first sufficient condition I know for a Borel probability measure $\mu$ (say) on $\mathbb{R}$ which has all moments of order $k$ for any $k\ge 1$ is as follows. If there exists $C>0$ such that for any $k\ge 1$, 
$$\int |x|^k d\mu(x)\le C R^k,$$
then 
$$\operatorname{supp}\mu \subseteq \{|x|\le R\}.$$
The proof uses Chebyshev's inequality. Let $\lambda>R$. Then,
$$\mu\left\{x\in \mathbb{R}:|x|>\lambda\right\}\le \frac{1}{\lambda^k}\int |x|^kd\mu(x)\le C \left(\frac{R}{\lambda}\right)^k\to 0$$
if we let $k\to\infty.$
Of course, the converse also holds.

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The second condition is the Paley-Wiener result given in this [answer][1].


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Unlike in the linked answers above, I am interested in merely *sufficient* conditions guaranteeing the compactness of the support with the hope that there are more results of this kind in addition to the ones I quoted above.

  [1]: https://mathoverflow.net/questions/129688/a-sufficient-condition-for-a-probability-measure-to-have-compact-support#:~:text=In%20order%20to%20avoid%20circularity,axis%20outside%20of%20this%20interval.